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Yakvenalex [24]
3 years ago
10

Could someone do all the problems and/or check my work?

Mathematics
2 answers:
MArishka [77]3 years ago
8 0
11. You’ve done it correctly

12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2

13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution

14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath

15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4

Hope this helped :)
erma4kov [3.2K]3 years ago
4 0

NOTES: To find the intercepts/roots:

  • move everything to one side and 0 on the other
  • factor the equation
  • apply the Zero Product Property (set each factor equal to 0)
  • solve for x

11. Answer: x = {-4, 0, 5}

<u>Step-by-step explanation:</u>

x^3=x^2+20x\\x^3-x^2-20x=0\\x(x^2-x-20)=0\\x(x-5)(x+4)=0\\x=0\quad x-5=0\quad x+4=0\\x=0\quad x=5\quad x=-4\\Solutions:\ x=\{-4, 0, 5\}

***********************************************************************************

12. Answer: x = No real solution

<u>Step-by-step explanation:</u>

x^4+13x^2+40=0\\(x^2+5)(x^2+8)=0\\x^2+5=0\quad \qquad x^2+8=0\\x^2=-5\qquad \qquad x^2=-8\\x=\pm \sqrt{-5}\quad \quad x=\pm \sqrt{-8}\\ \text{Neither solution is a real number}

***********************************************************************************

13. Answer: x = {-2, 2}

<u>Step-by-step explanation:</u>

x^4-x^2=x^2+8\\x^4-2x^2-8=0\\(x^2-4)(x^2+2)=0\\x^2-4=0\quad \qquad x^2+2=0\\x^2=4\qquad \qquad x^2=-2\\x=\pm \sqrt{4}\quad \quad x=\pm \sqrt{-2}\\x=\pm 2\qquad \text{x is not a real solution}\\Solutions: x = \{-2, 2\}

***********************************************************************************

14. Answer: \bold{x=\bigg\{-1, 1, -\dfrac{\sqrt{15}}{3}, \dfrac{\sqrt{15}}{3}\bigg\}}

<u>Step-by-step explanation:</u>

3x^4-8x^2+5=0\\3x^4-3x^2-5x^2+5=0\\3x^2(x^2-1)-5(x^2-1)=0\\(3x^2-5)(x^2-1)=0\\3x^2-5=0\qquad x^2-1=0\\x^2=\dfrac{5}{3}\qquad \qquad x^2=1\\\\x=\pm \sqrt{\dfrac{5}{3}}\quad \quad x=\pm \sqrt{1}\\\\x=\pm \dfrac{\sqrt{15}}{3}\qquad x=\pm1\\\\Solutions: x = \bigg\{-1, 1, -\dfrac{\sqrt{15}}{3}, \dfrac{\sqrt{15}}{3}}\bigg\}

***********************************************************************************

15. Answer: \bold{x=\bigg\{-2, 2, -\dfrac{5}{4}\bigg\}}

<u>Step-by-step explanation:</u>

4x^3+5x^2-16x-20=0\\3x^4-3x^2-5x^2+5=0\\x^2(4x+5)-4(4x+5)=0\\(x^2-4)(4x+5)=0\\x^2-4=0\qquad 4x+5=0\\x^2=4\qquad \qquad 4x=-5\\\\x=\pm \sqrt{4}\quad \quad x=\dfrac{-5}{4}\\\\x=\pm 2\qquad \quad x=-\dfrac{5}{4}\\\\ Solutions: x = \bigg\{-2, 2, -\dfrac{5}{4}}\bigg\}

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10 points someone help me please!?
Soloha48 [4]

Answer:

\large\boxed{b=\sqrt{95}}

Step-by-step explanation:

Use the Pyhagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=b,\ leg=7,\ hypotenuse=12

Substitute:

b^2+7^2=12^2

b^2+49=144       <em>subtract 49 from both sides</em>

b^2=95\to b=\sqrt{95}

7 0
3 years ago
The sum of an integer and 6 times the next consecutive odd integer is 61. Find the
vfiekz [6]

If the sum of an integer and 6 times the next consecutive integer is 61, the the value of lesser integer is 7

Consider the first odd integer as x

Then the next consecutive odd integer = x+2

The 6 times the second integer=  6(x+2)

= 6x+12

Sum of an integer and 6 times the next consecutive odd integer is 61

Then the equation will be

x + 6x+12 = 61

Add the like terms in the equation

(1+6)x + 12 = 61

7x +12 = 61

Move 12 to the right hand side of the equation

7x = 61-12

7x = 49

x = 49/7

x = 7

The second number is

x+2 = 7+2

= 9

Hence, if the sum of an integer and 6 times the next consecutive integer is 61, the the value of lesser integer is 7

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4 0
1 year ago
In the diagram below, C is the midpoint of AB. If AC is 4 centimeters, what is
Strike441 [17]

Answer:

The answer is B. 4cm

Step-by-step explanation:

Since it is the midpoint AC is always going to be the same as CB if C is the midpoint.

:)

6 0
2 years ago
A quadrilateral with two sets of parallel sides and four right angles is a
Musya8 [376]
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5 0
3 years ago
Read 2 more answers
Find the equation of a line through the coordinate (3,−2) and parallel to the line represented y=12x−2. A y=12x−2 B y=−2x+4 C y=
Hatshy [7]

Answer:

Step-by-step explanation:

y=12x−2. parallel lines have the same slope 12

find b at point (3,-2)

y=12x+b

-2=12(3)+b

-2-36=b

b=-36

the equation has to be y=12x-38

when you input the point(3,-2)

-2=12(3)-38

-2=-2 correct

7 0
3 years ago
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